On the concrete complexity of the successor function

Transcription

1 On the concrete complexity of the successor function M. Rigo joint work with V. Berthé, Ch. Frougny, J. Sakarovitch

2 Let s start with a quite naïve question. Just add one. Given the representation of the integer n, compute the representation of n +.

3 General framework Definition Let L be a language over a finite (totally) ordered alphabet (A,<). We order the words in L by increasing genealogical (or radix) order: w 0 w w 2 w n w n+ The successor function on L is Succ L : L L,w n w n+. Succ L (x) = y (x y) ( z L)((x z) ((y = z) (y z))).

4 Connection with abstract numeration systems An abstract numeration system is a triple S = (L,A,<) where L is an infinite regular language over the ordered alphabet (A, <) [P. Lecomte, M.R., 200]. Example (classical) Take L = {0,0} {ε}. ε We get back to the usual Zeckendorf numeration system based on the Fibonacci sequence.

5 Theorem [Ch. Frougny (997)] Let L be a regular language. The successor function Succ L is realized by a letter-to-letter transducer. Theorem [P.-Y. Angrand, J. Sakarovitch (200)] Let L be a regular language. The successor function Succ L is piecewise right sequential. sequential right transducer = co-sequential transducer transducer with deterministic underlying input automaton, reads and writes words from the right to the left. A function which is a finite union of (co-)sequential functions with pairwise disjoint domains is called a piecewise (co-)sequential function.

6 Theorem [M.-P. Schützenberger (975)] One can decide whether or not a transducer is functional (i.e., is realizing a rational function). Theorem [Ch. Choffrut (977)] One can decide whether or not a functional transducer is realizing a sequential function.

7 Theorem [P.-Y. Angrand, J. Sakarovitch (200)] A rational function is piecewise right sequential if and only if it can be realized by a cascade of sequential right transducers of height or 2.

8 Square of the Golden ratio X 2 3X +, β = , d β () = 2 ω, (U n ) n 0 =,3,8,2,... rep(n) = {ε,,2,0,,2,20,2,...} forbidden factors: 2 2; here Succ L is neither (left) sequential nor right sequential. P.-Y. Angrand [Ph.D. thesis (202), p. 28] (p) 2 22(q) 0000

9 One of the motivations stems from combinatorial, metrical, topological, dynamics, sequential properties of odometers A. M. Vershik, A theorem on the Markov periodical approximation in ergodic theory, J. Sov. Math. 28 (985) P. G. Grabner, P. Liardet, R. F. Tichy, Odometers and systems of numeration, Acta Arith. 70 (995) G. Barat, T. Downarowicz, P. Liardet, Dynamiques associées à une échelle de numération, Acta Arith. 03 (2002), Ch. Frougny,On-line odometers for two-sided symbolic dynamical systems, Proc. Lect. Notes in Comput. Sci (2002) V. Berthé, M. Rigo, Odometers on regular languages, Theory Comput. Syst. 40 (2007) 3.

10 We can compute, but how do we compute? Original problem (Words 2005) E. Barcucci, R. Pinzani, M. Poneti, Exhaustive generation of some regular languages by using numeration systems. For numeration systems built on some linear recurrent sequences of order 2, the amortized cost for computing rep(n +) from rep(n) is bounded by a constant (CAT).

11 Estimate the length of the carry propagation when applying the successor map on the first n words in L. amortized (or average) carry propagation. A computational issue: estimate the number of operations (in terms of Turing machines complexity) required to compute the representations of the first n integers from the first one by applying n times the successor function. (amortized) complexity, i.e., the average amount of computations required to obtain the successor of an element.

12 Framework We assume that L is right essential: L is prefix-closed; L is right extendable, w L, u ε : wu L. Example: Trie for the Zeckendorf system

13 Part I carry propagation (x,y) = { max( x, y ) if x y, min{ v u,w,x = uv,y = uw} if x = y. The carry propagation in the computation of Succ L (rep(i)) is (rep(i),rep(i +)). Example (Zeckendorf system) ε

14 Definition The (amortized) carry propagation of Succ L is defined as the following limit if it exists CP(Succ L ) = lim N N N i=0 (rep(i),rep(i +)). The limit might be infinite (e.g., language with polynomial growth, simple case: a ). The limit might not exist even for a right essential language.

15 u L (n) = #(L A n ) v L (n) = #(L A n ) Proposition Let L be a right essential language. The carry propagation for computing Succ L for all words of L of length n 0 is v L (n).

16 Let s have a look at the trie, v L (5) = 3. Fibonacci words of length

17 Let s have a look at the trie, v L (5) = 3. Fibonacci words of length

18 Let s have a look at the trie, v L (5) = 3. Fibonacci words of length

19 Let s have a look at the trie, v L (5) = 3. Fibonacci words of length

20 Let s have a look at the trie, v L (5) = 3. Fibonacci words of length

21 Theorem Let L be a right essential language. Suppose that lim u L(n +)/u L (n), or lim v L(n +)/v L (n), n n exists and equals some γ L >. Suppose that lim N N Then N i=0 (rep(i),rep(i +)) exists. CP(Succ L ) = γ L γ L. Can be applied to many classical numeration systems, for instance: trim minimal automaton M of L with a unique dominating eigenvalue γ L >, primitiveness of the trim minimal automaton, beta-numeration.

22 Part II Concrete complexity Suppose that P is a program (i.e., a Turing machine) which, for every i 0, computes Succ L (rep(i)) in Op(P,rep(i)) operations. The (amortized) complexity of P is comp(p) = lim N N N i=0 The (amortized) complexity of Succ L is Op(P,rep(i)). Comp(Succ L ) = inf{comp(p) P computes Succ L }. Remark If L is a regular language, then Comp(Succ L ) CP(Succ L ). At least the number of head moves corresponding to carry propagation.

23 Existence of a surcharge... Consider again the square of the Golden ratio : forbid 2 2 Succ(2) = CP = Comp = 2 Succ() = 2 CP = Comp The needed information to take a decision of move or writing. Carry propagation is not the only one that matters! CP(x) is the carry propagation, Comp(x) is the total number of operations needed to compute Succ L (x), The surcharge is the difference SC(x) = Comp(x) CP(x). The (amortized) surcharge for computing Succ L is SC(Succ L ) = lim N N N i=0 SC(rep(i)) = Comp(Succ L ) CP(Succ L ).

24 Proposition If Succ L is realized by a right sequential letter-to-letter finite transducer, then Comp(Succ L ) = CP(Succ L ). each move in the transducer is determined only by the input letter and produces an output letter, so there is no surcharge.

25 Case of beta-numeration β > real number Let d β () = (t n ) n be the (greedy) β-expansion of. = t i β i. i= If the β-expansion of is finite, d β () = t t m, then set v 0 =, v n = t v n + +t n v 0 + for n m, and v n = t v n + +t m v n m for n m. If the β-expansion of is infinite, d β () = (t n ) n, then set v 0 =, and v n = t v n + +t n v 0 + for n. The sequence V β = (v n ) n 0 with A β = {0,..., β } is the canonical numeration system associated with β. Note that lim n v n+ /v n = β.

26 If d β () is finite, β is a simple Parry number, or then if d β () is ult. periodic, β is a (non-simple) Parry number V β = (v n ) n 0 is a linear recurrent sequence; the language L(V) of the greedy expansions of all the non-negative integers is regular. Proposition [Ch. Frougny (997)] Let V be a linear recurrent sequence with dominant root β such that L(V) is regular. Then Succ L(V) is right sequential IFF (i) the β-expansion of is finite, of the form d β () = t t m, (ii) V is defined by v n = t v n + +t m v n m for n n 0 m and = v 0 < v < < v n0. Succ L(V) is right sequential, but cannot be realized by a letter-to-letter right sequential transducer whenever β N.

27 Sketch In the case of a simple Parry number, we can (algorithmically) build a specific right sequential transducer that computes Succ L(V) in such a way that we can derive a formula of the kind n SC(Succ Lβ ) = lim W i (src(e))δ(src(e))v n i (trg(e)). n v n e J i=0 where the quantities W l and V l refer to the number of particular paths of length l in the transducer and δ(e) is a known weight associated with particular edges in the construction.

28 Take the Pisot number β > being dominating root of the polynomial X 4 3X 3 3X 2 2X 2; dβ () = (332)ω. M = {ε,3,33,332,332,3323,33233,...} is the set of finite prefixes of dβ () = maximal representations for each length. 0, 0, 0, 2 0, 2 2, q Figure: Right automata L and M. Strategy : determine when the right-factor just being read is no more a maximal one.

29 M: in-splitting of M, (p, δ(p), λ(p)) = (state,valuation,label) Succ L ( 02) = 03, Succ L ( 02) = 022 but both reach state 3 q 0,0,ε 2 3 4,0,ε 3 2,, 3,,2 3 5,2,32 3 6,0,ε 2 3 3,2,2 3 5,3,32 Valuations give some information on the length of the right-factor read before detecting that a word does not belong to M.

30 The final transducer (needs outputs, edges of J in blue) 3 q 0,0,ε 4,0,ε ,, 3,,2 5,2,32 6,0,ε , 0, , 3 3,2,2 5,3,32 0, 2 2 0, ,3 3 0, , 0, 2 0, 0

31 Outputs on the edges The right-factor processed so far seems maximal: (A) in M, for each edge p a q with p s,q s where q is a terminal state, the corresponding edge in T is p a 0δ(p)+ q; (A2) in M, for each edge p a q where q is not a terminal state, the corresponding edge in T is p a ε q. We have just discovered that the right-factor is no more maximal: (A3) for each edge p a t in J such that p is a terminal state of M, there is an edge p a a+ t in T ; (A4) for each edge p a t in J such that p is not a terminal state of M, a a Succ(λ(p)) there is an edge p t in T. Nothing has to be done any more: a (A5) for each edge r t L the output is just the copy of the input, namely r a a t T.

32 For non-simple Parry number, we have developed a similar strategy with similar results about the amortized surcharge.